Question

Prove the following:

`(sin^2(-160^circ))/(sin^(2)70^circ) + sin(180^circ - theta)/sintheta` = sec²20°

Answer 1

sin(– 160°) = – sin 160°

= – sin(180° – 20°)

= – sin 20°

sin 70° = sin (90° – 20°) = cos 20°

∴ L.H.S. = `(sin^2(-160^circ))/(sin^(2)70^circ) + sin(180^circ - theta)/sintheta`

= `((-sin20^circ)^2)/((cos20^circ)^2) + sintheta/sintheta`

= `(sin^(2)20^circ)/(cos^(2)20^circ) + 1`

= 1 + tan²20°

= sec²20°

= R.H.S.